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Induction on a loop

  1. May 14, 2009 #1
    1. The problem statement, all variables and given/known data
    A 20.0 cm by 20.0 cm square loop of wire lies in the xy-plane with its bottom edge on the x-axis. The resistance of the loop is 0.520 {\Omega}. A magnetic field parallel to the z-axis is given by B = 0.790 * y^2 * t, where B is in tesla, y in meters, and t in seconds. What is the size of the induced current in the loop at t = 0.550 s?


    2. Relevant equations
    flux = A.B.cos theta


    3. The attempt at a solution

    I'm having a hard time understanding this question. I see that cos theta would be one so I can use the formula A B for the flux. But what is the y^2? Is it a variable? How can I construct the d(flux) / dt? I know that the A can be pulled out, and the 0.790 as well, but I'm not sure how to treat the other two.
     
  2. jcsd
  3. May 14, 2009 #2

    Redbelly98

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    Yes, y is a variable. Calculating the flux involves doing an integral.
     
  4. May 14, 2009 #3
    Okay but what variable is it? And why does calculating the flux involve doing an integral?
     
  5. May 14, 2009 #4

    Redbelly98

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    Do you understand what the xy plane is? x and y are just what they usually are. They are the x and y coordinates in the xy plane.

    The definition of flux is an integral:
    http://en.wikipedia.org/wiki/Magnetic_flux#Description
     
  6. May 14, 2009 #5
    Yes, somehow I can grasp what an x y plane is, but what I meant was for clarification was what it was the y coordinate of. Is it saying the field is different at different y coordinates. And the only definition of flux we have been given to use is area cross magnetic field.
     
    Last edited: May 14, 2009
  7. May 14, 2009 #6
    Yes, you are exactly right. However, since B is a function of y, you cannot just take out the area from the integral (the flux is not uniform over the entire square loop). You must divide the square into small area differentials dA.
     
  8. May 15, 2009 #7
    Yes, as tjkubo said, for this problem you can't simply say

    [tex]\Phi_B=BA\cos\theta[/tex]

    but instead must use

    [tex]\Phi_B=\int\vec{B}\cdot\textrm{d}\vec{A}[/tex]

    How would you write [itex]\textrm{d}\vec{A}[/itex] in terms of dy in this problem?
     
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